Question: Simplify the following expression and state the condition under which the simplification is valid. $y = \dfrac{4q^3 + 52q^2 + 120q}{3q^2 + 24q - 60}$
First factor out the greatest common factors in the numerator and in the denominator. $ y = \dfrac {4q(q^2 + 13q + 30)} {3(q^2 + 8q - 20)} $ $ y = \dfrac{4q}{3} \cdot \dfrac{q^2 + 13q + 30}{q^2 + 8q - 20} $ Next factor the numerator and denominator. $ y = \dfrac{4q}{3} \cdot \dfrac{(q + 10)(q + 3)}{(q + 10)(q - 2)}$ Assuming $q \neq -10$ , we can cancel the $q + 10$ $ y = \dfrac{4q}{3} \cdot \dfrac{q + 3}{q - 2}$ Therefore: $ y = \dfrac{ 4q(q + 3)}{ 3(q - 2)}$, $q \neq -10$